12 Oktober 2009

TUGAS 5 PDM

UNION

1. Show that (A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C)
Proof:
i) Show that (A ∪ B) ∩ C ⊂ (A ∩ C) ∪ (B ∩ C)
Take any x ∈ (A ∪ B) ∩ C
Obvious x ∈ (A ∪ B) ∩ C
≡ x ∈ (A ∪ B) ∧ x ∈ C
≡ (x ∈ A ∨ x ∈ B) ∧ x ∈ C
≡ (x ∈ A ∧ x ∈ C) ∨ (x ∈ B ∧ x ∈ C)
≡ x ∈ (A ∩ C) ∪ x ∈ (B ∩ C)
≡ x ∈ (A ∩ C) ∪ (B ∩ C) .......1)
We get for all x ∈ (A ∪ B) ∩ C then x ∈ (A ∩ C) ∪ (B ∩ C).
It means (A ∪ B) ∩ C ⊂ (A ∩ C) ∪ (B ∩ C)

ii) Show that (A ∩ C) ∪ (B ∩ C) ⊂ (A ∪ B) ∩ C
Take any x ∈ (A ∩ C) ∪ (B ∩ C)
Obvious x ∈ (A ∩ C) ∪ (B ∩ C)
≡ x ∈ (A ∩ C) ∨ (B ∩ C)
≡ (x ∈ A ∧ x ∈ C) ∨ (x ∈ B ∧ x ∈ C)
≡ (x ∈ A ∧ x ∈ C) ∨ (x ∈ B ∧ x ∈ C)
≡ x ∈ C ∧ (x ∈ A ∨ x ∈ B) (Asso)
≡ (x ∈ A ∨ x ∈ B) ∧ x ∈ C
≡ x ∈ (A ∪ B) ∩ C .......2)
From 1) and 2) we conclude that (A ∪ B) ∩ C is true.
So, (A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C)

TUGAS 4 PDM

2). b. Show that
A ∩ B = B ∩ A
Proof :
(i) Show that A ∩ B ⊂ B ∩ A
Take any x ∈ A ∩ B
Obvious x ∈ A ∩ B
≡ x ∈ A ∧ x ∈ B
≡ x ∈ B ∧ x ∈ A (komutatif)
≡ x ∈ (B ∩ A)
So, A ∩ B ⊂ B ∩ A .......(1)


(ii) Show that B ∩ A ⊂ A ∩ B
Take any x ∈ B ∩ A
Obvious x ∈ B ∩ A
≡ x ∈ B ∧ x ∈ A
≡ x ∈ A ∧ x ∈ B (komutatif)
≡ x ∈ (A ∩ B)
So, B ∩ A ⊂ A ∩ B .......(2)

So, from (1) and (2) we conclude that A ∩ B = B ∩ A

c. Show that
(A ∩ B) ∩ C = A ∩ (B ∩ C)
Proof :
(i) Show that (A ∩ B) ∩ C ⊂ A ∩ (B ∩ C)
Take any x ∈ (A ∩ B) ∩ C
Obvious x ∈ (A ∩ B) ∩ C
≡ x ∈ A ∧ x ∈ B ∧ x ∈ C
≡ x ∈ A ∧ x ∈ (B ∧ C) (assosiatif)
≡ x ∈ A ∧ x ∈ (B ∩ C)
So, (A ∩ B) ∩ C ⊂ A ∩ (B ∩ C)........ (1)

(ii) Show that (A ∩ B) ∩ C ⊂ A ∩ (B ∩ C)
Take any x ∈ A ∩ (B ∩ C)
Obvious x ∈ A ∩ (B ∩ C)
≡ x ∈ A ∧ x ∈ B x ∈ C
≡ x ∈ (A ∧ B) ∧ x ∈ C (assosiatif)
≡ x ∈ (A ∩ B) x ∈ C
So, A ∩ (B ∩ C) ⊂ (A ∩ B) ∩ C ........ (2)

So, from (1) and (2) we conclude that (A ∩ B) ∩ C = A ∩ (B ∩ C)